The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by :
Solution
NH<sub>2</sub>CONH<sub>2</sub> $\to$ NH<sub>3</sub>
<br><br>Using Principle of Atom Conservation
<br><br>2 $\times$ moles of urea = 1 $\times$ moles of NH<sub>3</sub>
<br><br>$\Rightarrow$ 2 $\times$ ${{0.6} \over {60}}$
<br><br>$\Rightarrow$ moles of NH<sub>3</sub> = 0.02
<br><br>Also moles of NH<sub>3</sub> = moles of HCl, because they react in 1 : 1 ratio.
<br><br>100 ml of 0.2 N HCl = ${{100 \times 0.2} \over {1000}}$ = 0.02 mole of HCl
<br><br>So option (B) is correct.
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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